A=(1+2+2^2)+2^3(1+2+2^2)+...+2^96(1+2+2^2)+2^99

=7(1+2^3+...+2^96)+2^99 ko chia hết cho 7

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

^A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

^A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

\(A=2^0+2^1+2^2+2^3+2^4+...+2^{99}=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)

A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ … + 2⁹⁹

A=( 2⁰ + 2¹ + 2² + 2³ + 2⁴⁾ +( 2⁵ … + 2⁹⁹⁾

A= 2⁰.(2⁰ + 2¹ + 2² + 2³ + 2⁴)+2⁵.( 2⁵ … + 2⁹⁹⁾

A= 1. 31+ 2⁵.31… + 2⁹⁵.31

A= 31. (1+2⁵...+2⁹⁵)

KL: ...... 

\(A=1+2+2^2+2^3+....+2^{98}+2^{99}\\ \Leftrightarrow A=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+....+\left(2^{98}+2^{99}\right)\\ \Leftrightarrow A=3+2^2.\left(1+2\right)+2^4.\left(1+2\right)+....+2^{98}.\left(1+2\right)\\ \Leftrightarrow A=3+3.2^2+3.2^4+....+3.2^{98}\\ \Leftrightarrow A=3.\left(1+2^2+2^4+...+2^{98}\right)⋮3\)

Ta có: \(A=2+2^2+2^3+2^4+...+2^{99}+91\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)+91\)

\(=2\cdot\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)+91\)

\(=7\cdot\left(1+2^4+...+2^{97}\right)+7\cdot13\)

\(=7\cdot\left(1+2^4+...+2^{97}+13\right)⋮7\)(đpcm)

Ta có: \(A=2+2^2+2^3+2^4+...+2^{99}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)\)

\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)

\(=\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{97}\right)\)

\(=7\cdot\left(2+2^4+...+2^{97}\right)⋮7\)(đpcm)

Ta có: A=2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰

-> A=2(1+2)+2³(1+2)+...+2⁹⁹(1+2)

->A=2.3+2³.3+...+2⁹⁹.3

->A=3(2+2³+...+2⁹⁹)\(⋮\)3

=> Đpcm

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)

\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)

\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31